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As the title states, I want to prove that if $G$ is a group with no proper subgroups and order at least 2, then $G$ is cyclic with prime order. This has been proven before in here, but I want to know if my own proof and explanations are correct.

Proof: Since $|G|>1$, we know that $\exists g∈G :|g|>1$. Consider the subgroup $⟨g⟩⊆G$ which is cyclic by definition. Since $G$ has no proper subgroups, we must have $⟨g⟩=G$. It is now to be proven that $|g|=p$ with $p=$ prime. I will show it in two steps, both by contradiction (edit: as pointed out, they are really proofs by contraposition): 1) $$ is finite. 2) $||=$ with $=$ prime.

  1. Assume that $|g|=\infty$. Now, $g^2\in\langle g\rangle$ so $\langle g^2\rangle$ is a subgroup of $\langle g\rangle$. Because $g^{n} \notin\left\langle g^{2}\right\rangle$ whenever $n$ is odd, $\langle g^2\rangle$ is a proper subgroup of $\langle g\rangle$. This contradicts the assumption that $G$ has no proper subgroups, so $g$ must be finite.
  2. Define the order of $g$ by $n$ and assume that $n$ is not prime, i.e., $\exists d\in \mathbb{Z} :d|n$ with $1<d<n$. Because $g^d\in\langle g\rangle$, we know that $\langle g^d\rangle $ is a subgroup of $\langle g\rangle$. Additionally, a theorem tells us that $|g^d|=n/d<n$ so $\langle g^d\rangle$ is a proper subgroup of $\langle g\rangle$. This is again a contradiction, so $n$ must be prime.

Is my proof correct and sufficient?

Logi
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    When you wrote "Assume that $g$ is infinite", you meant "Assume that $|g|$ is infinite". And in the second part, you meant to require that $1<d<n$. – lulu Apr 05 '22 at 18:23
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    In my book, "Abstract Algebra" by T. Hungerford, saying that $g$ is infinite means $|g|=\infty$, but maybe that is unconventional terminology. For the latter part, yes, I meant $1<d<n$. – Logi Apr 05 '22 at 18:27
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    Other than those small points, I'd say it looks good. – lulu Apr 05 '22 at 18:31
  • @lulu, thanks for looking it through. – Logi Apr 05 '22 at 19:00
  • You can actually prove the order is finite, and either odd or 2 without relying on an argument by contradiction by eliminating the superfluous assumption in your first case. Simply consider $\langle g^2\rangle$. If it is trivial, then $g$ has order 2 and you are done. Otherwise, it equals $\langle g\rangle$, so there exists $k$ such that $g=(g^2)^k$, so $g^{2k-1}=e$, showing the order is finite and divides $2k-1$ (hence is odd). Then use a similar argument to show that if $d$ divides the order of $g$, then it is either $1$ or the order, so the order is prime. (cont) – Arturo Magidin Apr 06 '22 at 00:51
  • (cont) this removes the unnecessary proofs-by-contradiction arguments which are really direct arguments with an unnecessary hypothesis at the beginning and a contradiction observed at the end. – Arturo Magidin Apr 06 '22 at 00:52

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Your argument is correct. I will point out that your argument in 1 is what I sometimes call a "proof by fake contradiction" or "fake proof by contradiction" (see discussion here). They are really proofs by contrapositive: you are showing that if the order $g$ is not a prime, then $G$ will not satisfy the hypotheses.

You can turn your argument into a direct proof that $g$ must have finite order:

Let $g\neq e$. Then $\langle g\rangle = G$. Now, $\langle g^2\rangle$ is either trivial (in which case $g$ has order $2$ and you are done); or else $\langle g^2\rangle = G = \langle g\rangle$. In this case, $g\in\langle g^2\rangle$, so there exists $k$ such that $g=(g^2)^k = g^{2k}$, hence $1=g^{2k-1}$ shows $g$ has finite order.

The second part can likewise be done directly: let $n=|g|$, which we now know is finite. If $1\lt d\lt n$, then $\langle g^d\rangle = \langle g\rangle$, so there exists $k$ such that $g=g^{dk}$. Then $n\mid dk-1$, so $\gcd(n,d)=1$. Thus, $n$ is relative prime to all integers $d$ with $1\lt d\lt n$, so we conclude that $n$ is prime, as required.

Arturo Magidin
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  • Thanks for your input. You are correct, the two proofs are both proofs by contraposition; in the first I take "$|g|$ is not finite" as the premise and shows that it follows that $\langle g\rangle$ has proper subgroups. In the following proof I take as premise that the order of $|g|$ is not prime, and show that it must follow that $\langle g\rangle$ has proper subgroups. But a proof by contraposition is a valid proof, so why do you suggest that I change it to a direct proof? – Logi Apr 06 '22 at 19:19
  • @Logi: The fact that you cast them as proofs by contradiction suggested to me that you perhaps wanted a direct proof; there is no problem with a proof by contrapositive, and if you'd written it as a proof by contrapositive explicitly then I would not have made a comment. Also, it's a personal pet peeve about people who write direct proofs/proofs by contrapositive, but write them as proofs by contradiction when there is no need to do so. – Arturo Magidin Apr 06 '22 at 19:27
  • I just wanted a valid proof, but yes I mixed up the terminology (I will remember not to denote a proof by contraposition as a proof by contradiction :)) – Logi Apr 06 '22 at 19:41